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12 June, 23:06

determine the magnitude and units of the constant k. What is the net displacement of the particle over the same 6-second interval of motion?

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  1. 13 June, 02:07
    0
    54 m

    Explanation:

    Compute v (t):

    a = dv / dt = - k*t^2

    dv = - k*t^2. dt

    Integrating both sides:

    v = - kt^3 / 3 + C

    v (0) = 12

    Hence,

    v (t) = 12 - k*t^3 / 3

    Compute k:

    @ t = 6s v (t) = 0

    12 - k*t^3 / 3 = v (t)

    12 - k * (6) ^3 / 3 = 0

    k = 12 / 72 = 1 / 6 m/s^2

    Compute s (t)

    v = ds / dt = 12 - t^3 / 18

    ds = (12 - t^3 / 18). dt

    Integrating both sides

    s (t) = s (0) + (12*t - t^4 / 72)

    Net displacement:

    s (6) - s (0) = 12*t - t^4 / 72

    = 12*6 - 6^4 / 72

    = 54 m
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